It is given that $(t-1)^3$ is a solution of $(t-1)^2y''-(t-1)y'-3y=0$ Find a general solution of the equation. It is given that $(t-1)^3$ is a solution of $(t-1)^2y''-(t-1)y'-3y=0$ Find a general solution of the equation.
I set $z=t-1$ so the equation becomes $z^2y''-zy'-3y=0$ which is a Euler equation.
Then I get the auxiliary eq: $m^2-2m-3=0 \implies m=3,m=-1$
Hence the general solution will be $y=c_1z^3+c_2z^{-1}$ and if I substitute then I will get $y=c_1(t-1)^3+{c_2 \over t-1}$
So I am wondering what is the use of the given data that $(t-1)^3$ is a solution?? Am I missing something?
 A: Like MrYouMath stated in his comment, if you have $n$ independent solutions to a $m$th order differential equation, you can reduce the differential equation by $n$ orders.
The method which allows you to do this is equating a new solution to the differential equation to the product of a unknown function and a known solution. So in this case
$$
y_1(t) = (t-1)^3
$$
$$
y_2(t) = v(t) y_1(t).
$$
By using the product rule, you find expressions, in terms of $y_1(t)$ and $v(t)$, for the first and second order derivative of $y_2(t)$. By substituting these expressions in the differential equation you will see that all terms which are multiplied by $v(t)$ cancel, because $y_1(t)$ is a solution. This means that you will only be left with first and second order derivative of $v(t)$ and thus you are left with a first order differential equation of $w(t)$, defined as $v'(t)$. Integrating $w(t)$ will give you $v(t)$, but also adds a constant, which can be used to solve a boundary value problem.
Note that your approach of finding all solutions to the initial differential equation does also work, but reduction of order can help simplify the problem, which might make it easier to solve.
